Problem: Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $n \neq 0$. $r = \dfrac{n^2 + 10n}{n^3 + 9n^2 - 10n} \times \dfrac{4n^2 + 12n - 16}{n - 1} $
Explanation: First factor out any common factors. $r = \dfrac{n(n + 10)}{n(n^2 + 9n - 10)} \times \dfrac{4(n^2 + 3n - 4)}{n - 1} $ Then factor the quadratic expressions. $r = \dfrac {n(n + 10)} {n(n - 1)(n + 10)} \times \dfrac {4(n - 1)(n + 4)} {n - 1} $ Then multiply the two numerators and multiply the two denominators. $r = \dfrac {n(n + 10) \times 4(n - 1)(n + 4) } { n(n - 1)(n + 10) \times (n - 1)} $ $r = \dfrac {4n(n - 1)(n + 4)(n + 10)} {n(n - 1)(n + 10)(n - 1)} $ Notice that $(n - 1)$ and $(n + 10)$ appear in both the numerator and denominator so we can cancel them. $r = \dfrac {4n\cancel{(n - 1)}(n + 4)(n + 10)} {n\cancel{(n - 1)}(n + 10)(n - 1)} $ We are dividing by $n - 1$ , so $n - 1 \neq 0$ Therefore, $n \neq 1$ $r = \dfrac {4n\cancel{(n - 1)}(n + 4)\cancel{(n + 10)}} {n\cancel{(n - 1)}\cancel{(n + 10)}(n - 1)} $ We are dividing by $n + 10$ , so $n + 10 \neq 0$ Therefore, $n \neq -10$ $r = \dfrac {4n(n + 4)} {n(n - 1)} $ $ r = \dfrac{4(n + 4)}{n - 1}; n \neq 1; n \neq -10 $